Re: [Haskell-cafe] eager/strict eval katas

Thomas Hartman wrote:

...

Note that 1 + ··· + n = n * (n+1) / 2, so the average of [1..n] is (n+1) / 2

fair enough.

But I believe if I restate the problem so that you need to find the average of an arbitrary list, your clever trick doesn't work and we need eager eval or we blow the stack.

Not true: Prelude Data.List> let f a = (\(a,b,c)->c) . head . dropWhile (\(s,n,_) -> s <=n*a) . scanl (\(s,n,_) x ->(s+x,n+1,x)) (0,0,0) in f (10^5) [1,3..] 200001

Also... on second thought, I actually solved a slightly different problem than what I originally said: the problem of detecting when the moving average of an increasing list is greater than 10^6; but my solution doesn't give the index of the list element that bumped the list over the average. However I suspect my code could be tweaked to do that (still playing around with it):

Also I actually used a strict scan not a strict fold and... ach, oh well.

scanl above is not strict in its second argument. The data dependencies cause the strictness. Cf: Prelude> head ([1,3] ++ head ((scanl undefined undefined) undefined)) 1

As you see I wrote a customized version of foldl' that is strict on the tuple for this to work. I don't think this is necessarily faster than what you did (haven't quite grokked your use of unfold), but it does have the nice property of doing everything in one one fold step (or one scan step I guess, but isn't a scan

http://thomashartman-learning.googlecode.com/svn/trunk/haskell/lazy-n-strict...

You have Prelude Control.Arrow Data.List> let avg5 = uncurry (/) . foldl' (\(s,n) x -> (s + x,n + 1)) (0,0) in avg5 [1..10000000] *** Exception: stack overflow -- This fails in 100 sec Try this. It is not foldl' that needs to be strict, but the function folded: Prelude Data.List> let avg5 = uncurry (/) . foldl' (\(!s,!n) x -> (s + x,n + 1)) (0,0) in avg5 [1..10000000] You will need -fbang-patterns for this (there are other ways to do this in Haskell 98 though).

t.

t1 = average_greater_than (10^7) [1..]

average_greater_than max xs = find (>max) $ averages xs

averages = map fst . myscanl' lAccumAvg (0,0) average = fst . myfoldl' lAccumAvg (0,0) lAccumAvg (!avg,!n) r = ( (avg*n/n1) + (r/n1),(n1)) where n1 = n+1

myfoldl' f (!l,!r) [] = (l,r) myfoldl' f (!l,!r) (x:xs) = ( myfoldl' f q xs ) where q = (l,r) `f` x

myscanl f z [] = z : [] myscanl f z (x:xs) = z : myscanl f (f z x) xs

myscanl' f (!l,!r) [] = (l,r) : [] myscanl' f (!l,!r) (x:xs) = (l,r) : myscanl' f q xs where q = (l,r) `f` x

*"Felipe Lessa" *

12/12/2007 02:24 PM

To Thomas Hartman/ext/dbcom@DBAmericas cc haskell-cafe@haskell.org Subject Re: [Haskell-cafe] eager/strict eval katas

On Dec 12, 2007 2:31 PM, Thomas Hartman wrote:

...

exercise 2) find the first integer such that average of [1..n] is > [10^6] (solution involves building an accum list of (average,listLength) tuples. again you can't do a naive fold due to stack overflow, but in this case even strict foldl' from data.list isn't "strict enough", I had to define my own custom fold to be strict on the tuples.)

What is wrong with

Prelude> snd . head $ dropWhile ((< 10^6) . fst) [((n+1) / 2, n) | n <- [1..]] 1999999.0

Note that 1 + ··· + n = n * (n+1) / 2, so the average of [1..n] is (n+1) / 2. The naive

Prelude Data.List> let avg xs = foldl' (+) 0 xs / (fromIntegral $ length xs) Prelude Data.List> snd . head $ dropWhile ((< 10^6) . fst) [(avg [1..n], n) | n <- [1..]]

works for me as well, only terribly slower (of course). Note that I used foldl' for sum assuming the exercise 1 was already done =). How did you solve this problem with a fold? I see you can use unfoldr:

Prelude Data.List> last $ unfoldr (\(x,s,k) -> if s >= k then Nothing else Just (x, (x+1,s+x,k+10^6))) (2,1,10^6)

I'm thinking about a way of folding [1..], but this can't be a left fold (or else it would never stop), nor can it be a right fold (or else we wouldn't get the sums already done). What am I missing?

Cheers,

-- Felipe.

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